Why $L^2(\Omega) \hookrightarrow H^{-1}$?

Question

After the characterisation theorem for a $F \in H^{-1}(\Omega)$, i.e.

$H^{-1}$ is the set of distributions such that $F= f_0 + \operatorname{div}(\boldsymbol{f})$, with $f_0 \in L^2(\Omega)$ and $\boldsymbol{f}=(f_1,\ldots,f_n)$ a vector of $L^2$ functions. Also, $$||F||_{H^{-1}} \leq C_P ||f_0||_0 + ||\boldsymbol{f}||_0$$

my professor wrote that $$L^2(\Omega) \hookrightarrow H^{-1}(\Omega)$$ I don't know if I can show this, and especially how I should move.

I think I should prove that $$||F||_{H^{-1}} \leq ||F||_{L^2}$$ where both norms are dual norms.

Here's what I tried:

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Now, $$||F||_{H^{-1}} = \sup \{|Fv|, v\in H^1_0, ||v||_1 \leq 1  \}$$ and $$||F||_{L^{2}} = \sup \{|Fv|, v\in L^2, ||v||_0 \leq 1  \}$$ Since $H_0^1 (\Omega) \subset H^1(\Omega) \subset L^2(\Omega)$

then in the $||F||_{L^2}$ I'm taking the supremum over a larger set, so it should be bigger than $||F||_{H^{-1}}$, right?

Answer 1

I think you're overthinking things. By your characterization, taking $F = f_0$ you know that $\lVert F \rVert_{H^{-1}} = \lVert f_0 \rVert_{H^{-1}} \leq C_P\lVert f_0 \rVert_0$. This exactly means that the inclusion $L^2(\Omega) \hookrightarrow H^{-1}(\Omega)$ is bounded with norm at most $C_P$.